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Volume of a Wigner Sietz cell

  1. Feb 24, 2010 #1
    I have a few questions, first of all I'm trying to figure out how to find the volume of a 3 dimmensional wigner sietz cell. I have the 8th edition intro to solid state physics book by charles kitle and there is no where in the book that shows me how to find the volume of wigner seitz cell.

    The question that I am trying to get at is there is a (110) plane of a FCC lattice, covering 4 conventional cells in a 2x2 array. If the side length of the conventional cell is 3.7 Angstrums, evaluate the nearest neighbor distance d("hard sphere" diameter) and the volume of the Wigner-Seitz cell.

    First of all, what I don't understand is I thought the Wigner Seitz cell could be stacked perfectly and so I thought that the neighbors line up and touch each other perfectly like stacking cubes. Also could you explain what a "hard sphere" diameter is because no where in my book or ANY online sources can I find what a "hard sphere" diameter is. The second problem was, find the volume of the Wigner Seitz cell and so I thought that by making a 3d cart. coord graph and creating planes that intersect at the normal of the lines at the half way points i could find the volume, but unfortunately no matter what I do it seems to complicated for me. Is their a simple formula or something i could use?

    Lastly I am also asked to used the hard sphere diameter to evaluate the distance from an atom to its closest next nearest neighbor in the second layer above it for both the FCC and HCP stacking. The seperation between those planes is given by root(8/3). I have no idea what the god my proffessor is talking about, I'm having a heart attack and I really want to understand the material, but as I said none of this stuff is in my book nor in any online lectures or wikipedia or whatever else is out there.

    Thanks for your time and replies.
     
  2. jcsd
  3. Feb 25, 2010 #2
    The Wigner-Seitz cell is space filling. So you just need to find some volume that you can calculate easily, and count how many Wigner-Seitz cells are inside. For an fcc lattice, the conventional cell is just a cube of side a with 4 atoms in it, so the Wigner-Seitz cell has a volume of a^3/4.

    The "hard sphere" diameter is what you would have if you had a lattice of marbles that just touch each other. They are hard because they don't overlap or anything. It's just a nice classical physics picture for how atoms arrange in a lattice.

    There's a post 10-15 messages down about calculating the c/a ratio for HCP. Read through that one.
     
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